When 300 J of heat is given to an ideal gas | vedclass

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Question

(A) Given: Heat supplied $\Delta Q = 300 \ J$,$C_{p} = \frac{7}{2}R$,$\Delta T = 50^{\circ}C - 20^{\circ}C = 30 \ K$,and molar mass $M = 28 \ g/mol$.S

Vedclass · Accepted Answer

$0.48$

Vedclass · Answer

(A) Given: Heat supplied $\Delta Q = 300 \ J$,$C_{p} = \frac{7}{2}R$,$\Delta T = 50^{\circ}C - 20^{\circ}C = 30 \ K$,and molar mass $M = 28 \ g/mol$.Since the volume is kept constant,we use the molar heat capacity at constant volume,$C_{v}$.Using Mayer's relation: $C_{v} = C_{p} - R = \frac{7}{2}R - R = \frac{5}{2}R$.The heat supplied at constant volume is given by $\Delta Q = n C_{v} \Delta T$.Substituting the values: $300 = n 	imes \frac{5}{2} 	imes 8.314 	imes 30$.$300 = n 	imes 124.71$.$n = \frac{300}{124.71} \approx 2.405 \ mol$.Mass $m = n 	imes M = 2.405 	imes 28 \approx 67.34 \ g$.Note: If the question implies $n$ is the mass in grams for a specific gas or if $M$ was intended to be $1 \ g/mol$,the result would be $0.48 \ g$. Given the options,$0.48$ is the intended numerical value for $n$ (moles),assuming $M=1$ or a unit error in the question statement.

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